# Problem proving the last steps of Birkhoff's ergodic theorem

Birkhoff's ergodic theorem states: Let $$(\Omega, \mathbb{F}, P, T)$$ an ergodic system, i.e, $$(\Omega, \mathbb{F}, P)$$ is a probability space and and $$T:\Omega \to \Omega$$ such that: $$\forall A \in \mathbb{F}$$, $$P(T^{-1}(A))=P(A)$$. Moreover, $$A=T^{1}(A) \implies P(A) \in \{0,1\}$$.

Now, let $$X \in L^{1}(\Omega,P)$$. Then $$\frac{1}{n} \sum_{k=0}^{n-1} X \circ T^{k} = \mathbb{E}(X)$$ a.s, i.e $$P(\{w: \frac{1}{n} \sum_{k=0}^{n-1} X \circ T^{k}(w) = \mathbb{E}(X)\})=1$$.

We now define the following: Let $$\epsilon>0$$ and $$f\in L^{1}(\Omega, P)$$

$$Sn(f)=\sum_{k=0}^{n-1} f \circ T^{k}$$

$$g_{\epsilon}=X-\mathbb{E}(X)-\epsilon$$

$$G_{\epsilon} = \limsup \frac{1}{n}Sn(g_{\epsilon})$$

We prove this theorem by proving the following steps:

1. $$G_{\epsilon} \circ T = G_{\epsilon}$$ and $$P(\{G_{\epsilon}>0\}) \in \{0,1\}$$
2. Let $$M_{n}=max\{0,S_{1}(g_{\epsilon}),S_{2}(g_{\epsilon}),...,S_{n}(g_{\epsilon})\}$$. We prove can prove that for $$w\in \{M_{n}>0\}$$, $$g_{\epsilon}\geq M_{n}(w)-M_{n}(T(w))$$, so we can conclude that $$\mathbb{E}(g_{\epsilon}\mathbb{1}_{M_{n}>0})\geq 0$$
3. Let $$A= \bigcup_{n=1}^{\infty} \{S_{n}(g_{\epsilon})>0\}$$, then $$\mathbb{E}(g_{\epsilon}\mathbb{1}_{A})\geq 0$$
4. Conclude that $$P( \limsup \frac{1}{n}S_{n}(X) \leq \mathbb{E}(X))=1$$ for any $$X \in L^{1}(\Omega,P)$$
5. Conclude that $$P( \limsup \frac{1}{n}S_{n}(X) = \mathbb{E}(X))=1$$ for any $$X \in L^{1}(\Omega,P)$$

MY PROBLEM:

So far I've proven 1,2,3 using these techniques:

1. Just manipulating the terms and applying some properties
2. This is Hopf maximal ergodic lemma
3. Applying DCT

Now, Im totally lost on how to prove 4 and 5 which is the conclusion. Any help would be enormously appreciated.

Thanks so much for your help <3!

• $\text{\limsup}$ is a thing in latex Commented Oct 2, 2019 at 15:17
• $Sn(X)=\frac{1}{n} \sum_{k=0}^{n-1} X \circ T^{k}$ Commented Oct 2, 2019 at 15:19
• $S_n(x)$? ${}{}$ Commented Oct 2, 2019 at 15:20
• Look at step 2 ${}$ Commented Oct 2, 2019 at 15:23
• I don't think the factors of $\frac{1}{n}$ should be appearing in statements 4 and 5, since it's already in the definition of $S_n$. Commented Oct 2, 2019 at 15:40

I don't think you can prove (4) based on just (1),(2), and (3). What you want instead of (3) is $$\mathbb{E}(g_\epsilon1_B) \ge 0$$ for each $$T$$-invariant $$B\subseteq A$$. In the below, I assume $$\mathbb{E}(X) = 0$$ (just replace $$X$$ with $$X-\mathbb{E}(X)$$).
If we have this improved (3), then for any $$\beta < \epsilon$$, $$B_\beta = \{\omega \in \Omega: \limsup_n \frac{1}{n}\sum_{k=0}^{n-1} X(T^k\omega) > \epsilon \text{ and } \liminf_n \frac{1}{n}\sum_{k=0}^{n-1} X(T^k\omega) < \beta\}$$, which is certainly $$T$$-invariant and a subset of $$A$$. We thus get $$\int_{B_\beta} X \ge \epsilon|B_\beta|$$. Now, replacing $$X$$ with $$-X$$, $$\epsilon$$ with $$-\beta$$, and $$\beta$$ with $$-\epsilon$$ leaves $$B_\beta$$ unchanged, and so doing everything for $$-X$$ gives $$\int_{B_\beta} (-X) \le -\beta|B_\beta|$$, i.e., $$\int_{B_\beta} X \ge \beta|B_\beta|$$. Since $$\beta < \epsilon$$, we conclude $$|B_\beta| = 0$$. Taking $$\beta$$ arbitrarily close to $$\epsilon$$, we see $$\mathbb{P}(\{\omega : \Omega : \limsup_n \frac{1}{n}\sum_{k=0}^{n-1} X(T^k\omega) > \epsilon\}) = 0$$. Now just take $$\epsilon$$ arbitrarily close to $$0$$ to get (4). Doing everything for $$-X$$ then gives (5).